How can we integrate:
$$ \int\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}dx $$
Using simple algebraic identities I deduced it to
$$ \int\frac{1-2\sin^2x\cdot\cos^2x}{(\sin x+\cos x)(1-\sin x\cdot\cos x)}dx $$ but can't proceed further. Please provide some directions?
$\displaystyle\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}=\sin x+\cos x-\frac{\sin x\cos x}{\sin^3 x+\cos^3x}=\sin x+\cos x-\frac{\sin x\cos x}{(\sin x+\cos x)(1-\sin x\cos x)}$,
and $\;\;\displaystyle\frac{\sin x\cos x}{(\sin x+\cos x)(1-\sin x\cos x)}=A\left(\frac{\sin x+\cos x}{1-\sin x\cos x}\right)+B\left(\frac{1}{\sin x+\cos x}\right)$
where $\sin x\cos x=A(\sin x+\cos x)^2+B(1-\sin x\cos x)=(A+B)+(2A-B)\sin x\cos x$.
Then $A=\frac{1}{3}$ and $B=-\frac{1}{3}$,
so $\displaystyle\int\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}dx=\int\left(\sin x+\cos x-\frac{1}{3}\cdot\frac{\sin x+\cos x}{1-\sin x\cos x}+\frac{1}{3}\cdot\frac{1}{\sin x+\cos x}\right)dx$
$\displaystyle=-\cos x+\sin x-\frac{1}{3}\int\frac{2\sin x+2\cos x}{2-2\sin x\cos x}dx+\frac{1}{3}\int\frac{1}{\sqrt{2}\sin(x+\frac{\pi}{4})}dx$
$\displaystyle=-\cos x+\sin x-\frac{2}{3}\int\frac{\sin x+\cos x}{1+(\sin x-\cos x)^2}dx+\frac{1}{3\sqrt{2}}\int\csc\left(x+\frac{\pi}{4}\right)dx$
$\displaystyle=-\cos x+\sin x-\frac{2}{3}\arctan(\sin x-\cos x)+\frac{1}{3\sqrt{2}}\ln\big|\csc\left(x+\frac{\pi}{4}\right)-\cot\left(x+\frac{\pi}{4}\right)\big|+C$
$\displaystyle=-\cos x+\sin x-\frac{2}{3}\arctan(\sin x-\cos x)+\frac{1}{3\sqrt{2}}\ln\left\vert\frac{\sqrt{2}-\cos x+\sin x}{\sin x+\cos x}\right\vert+C$